Since the assertion follows from Lemma REF  and REF  for \(e=n\) 
or \(e=n-1\) , respectively, in order to prove the theorem, it remains to
tackle the case \(e=n-2\)  and \(X\cong P\Omega _{n}^{+}(q)\) . Clearly, \(\left( \Phi _{(n-2)f}^{\ast }(p),\left|X_{x}\right|\right) >1\) .
Suppose that \(X_{x}\)  is not a large subgroup of \(X\) . Then \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\)  by
Lemma REF (2). Recall that \(n\geqslant 6\)  by Remark REF , and that \(\left|\mathrm {Out(}X\mathrm {)}\right|\)  is either \(2(4,q^{n/2}-1)f\) 
or \(6(4,q^{n/2}-1)f\)  according as \(n\ne 8\)  or \(n=8\)  respectively.
Since \(X_{x}<X<PSL_{n}(q)\)  and \(\Phi _{(n-2)f}^{\ast }(p)\)  divides \(\left|X_{x}\right|\) , the group \(X_{x}^{^}\)  is one the
groups listed in Theorem 3.1 of [1]}.
Assume that \(X_{x}^{^}\)  is a nearly simple subgroup of \(SL_{n}(q)\) . Suppose that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is alternating. If \(X_{x}^{^}\)  arises from the action of \(A_{\ell }\)  on its fully deleted
permutation module, then \(q\)  is a prime (hence \(f=1\) ), \(n=e+2\) , \(A_{e+4}\leqslant X_{x}\leqslant S_{e+4}\times Z_{2}\)  where \((p,e)=(2,4),(2,10),(2,12),(2,18),(3,4),(3,6),(5,6)\) . If \((p,e)\ne (3,4)\) ,
then \(\left|X_{x}\right|_{2}\geqslant 2^{6}\) , \(\left|X_{x}\right|_{3}\geqslant 9\)  and \(\left|X_{x}\right|_{5}\geqslant 25\) ,
whereas \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 24\)  in the
first case and \(\left|\mathrm {Out(}X\mathrm {)}\right|\leqslant 8\)  in
the remaining ones. So, these cases are excluded. Then \((p,e)=(3,4)\) , \(X\cong P\Omega _{8}^{+}(3)\)  and \(A_{8}\leqslant X_{x}\leqslant S_{8}\times Z_{2}\) .
However, this is impossible since 13 must divide \(\left|X_{x}\right|\) , since \(P\Omega _{8}^{+}(3)\)  is divisible by 13, and not by \(13^{2}\) , and since \(\left[ X:X_{x}\right] =k^{2}\) .
Suppose that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  does not arise from the action of \(A_{\ell }\)  on its
fully deleted permutation module. Then \(n=8\)  and either \(q=5\)  and \(\ell =7,8,9,10\) , or \(q=3\)  and \(\ell =9,8\) . The above argument excludes \(q=3\) . Case \(q=5\)  is ruled out too. Indeed, \(P\Omega _{8}^{+}(5)\)  is
divisible by 31 and not by \(31^{2}\) .
Assume that \(\mathrm {Soc(}X_{x}\mathrm {)}\)  is either a sporadic group or a simple Lie type group in cross
characteristic. Set \(S\cong \mathrm {Soc(}X_{x}\mathrm {)}\) . By Theorem 3.1 of
[1]}, one of the following holds:

\((n,q)=(6,3)\)  and \(S\cong M_{12},PSL_{3}(4)\) ;

\((n,q)=(8,3)\)  and \(S\cong P\Omega _{8}^{+}(2)\) , \(PSp_{6}(2)\) ;

\((n,q)=(8,5)\)  and \(S\cong PSL_{3}(4)\) , \(P\Omega _{8}^{+}(2)\) , \(PSp_{6}(2)\) , or \(Sz(8)\) ;

\((n,q)=(14,2)\)  and \(S\cong PSL_{2}(13),\)  \(G_{2}(3)\) ;

\((n,q)=(20,2)\)  and \(S\cong J_{2},PSL_{2}(19)\) .

\(M_{12},PSL_{3}(4)\)  are not subgroups of \(P\Omega _{6}^{+}(3)\) , hence Case
(1) cannot occur. Cases (2) and (3) cannot occur by Lemma REF . Finally, 127 divides the order of \(P\Omega _{14}^{+}(2)\)  and of \(P\Omega _{20}^{+}(2)\) . Then \(127^{2}\)  must divide the order of
\(S\) , since \(\left[ X:X_{x}\right] =k^{2}\) , and since \(S\trianglelefteq X_{x}\leqslant \mathrm {Aut(}S\mathrm {)}\) . However, this is impossible for \(S\)  as
in Cases (4) or (5), and hence these are ruled out.
The group \(X_{x}\)  cannot be a Classical Example by [3]}, Section 4.8, as \(X_{x}<X\) . The same
argument used above to rule out Case (1) excludes the possibility for \(X_{x}\)  to
be a Symplectic Type Example as well. Now, assume that is an Imprimitive
Example. Then \(q=p\) , \(e=n-2\) , \(\Phi _{e}^{\ast }(p)=e\)  and \(X_{x}^{^}\leqslant GL_{1}(q)\wr S_{n}\)  with \((n,q)=(6,2),(12,2),(14,2),(20,2),(6,3),(8,3),(8,5)\) . Actually, \((n,q)\ne (6,2)\)  since \(P\Omega _{6}^{+}(2)\cong A_{8}\)  has been ruled out in Lemma
3.5 of [4]}. In the remaining cases, there is a prime \(w\)  such that \(w\mid \left|X\right|\) , \(w^{2}\nmid \left|X\right|\) . Such
a prime \(w\)  is 31, 127, 257, 13, 13, 31, respectively. Then \(w\mid \left|X_{x}\right|\) , since \(\left[ X:X_{x}\right] =k^{2}\) ,
whereas \(w>n\)  and \(w \nmid q-1\) . So, \(X_{x}\)  cannot be an Imprimitive Example. Thus, again by
Theorem 3.1 of [1]}, either \(X_{x}\)  preserves a point or a hyperplane of \(PG_{n-1}(q)\) , or \(X_{x}^{^}\)  is an Extension Field Example.
Assume that \(X_{x}^{^}\)  is an Extension Field Example. Since \(X_{x}^{^}\)  preserves an orthogonal form on \(V_{n}(q)\) , Theorem
3.1 of [1]} implies that either \(n\equiv 2 \pmod {4}\) , \(q\)  is odd and \(X_{x}^{^}\leqslant \Gamma O_{n/2}(q^{2})\) , or \(n\equiv 0 \pmod {4}\) 
and \(X_{x}^{^}\leqslant \Gamma U_{n/2-1}(q)\)  respectively. Thus \(X_{x}\)  lies in a
maximal member \(Y\)  of \(\mathcal {C}_{3}(X)\)  of type \(O_{n/2}(q^{2})\)  or \(GU_{n/2-1}(q)\)  respectively. If \(X_{x}<Y\) , then \(X_{x}\)  is a type 1 novelty with respect
to \(Y\)  by Lemma REF (2). However, this is impossible by [3]},
Tables 3.5.H–I, for \(n\geqslant 13\)  and [8]} for \(6\leqslant n\leqslant 12\) . Therefore, \(X_{x}\)  is a maximal \(\mathcal {C}_{3}\) -subgroup of \(X\) . Thus, \(\left|X_{x}\right|_{p}\geqslant q^{2}\)  for \(n\ne 8\) , and \(\left|X_{x}\right|_{p}\geqslant q^{3}\)  for \(n=8\) . Hence, \(\left|X_{x}\right|_{p}<\left|\mathrm {Out(}X\mathrm {)}\right|\) 
implies \(n=8\)  and \(q=2\) , as \(\left|\mathrm {Out(}X\mathrm {)}\right|\) 
is either \(2(4,q^{n/2}-1)f\)  or \(6(4,q^{n/2}-1)f\)  according as \(n\ne 8\) , or \(n=8\) , respectively. However, \(X\ncong P\Omega _{8}^{+}(2)\)  by Lemma REF .
Finally, assume that \(X_{x}\)  preserves either a point or a hyperplane of \(PG_{n-1}(q)\) . Note that \(X_{x}\)  does not lie in a maximal parabolic subgroup of \(X\)  by Proposition REF . Then \(X_{x}\)  preserves either a non-degenerate
point, or a non-degenerate line of \(PG_{n-1}(q)\)  of type \(-\)  by [3]}, Proposition
4.1.6.(II), since \(\Phi _{n-2}^{\ast }(p)\mid \left|X_{x}\right|\) . So does \(G_{x}\) . Then \(X\cong P\Omega _{6}^{+}(q)\)  by [10]}, Lemma 3.13. However, this is impossible by Lemma REF , since \(X\cong PSL_{4}(q)\) . This completes the proof.
